Question: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{5n^2 - 55n + 120}{-5n^3 + 30n^2 - 45n}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {5(n^2 - 11n + 24)} {-5n(n^2 - 6n + 9)} $ $ r = -\dfrac{5}{5n} \cdot \dfrac{n^2 - 11n + 24}{n^2 - 6n + 9} $ Simplify: $ r = - \dfrac{1}{n} \cdot \dfrac{n^2 - 11n + 24}{n^2 - 6n + 9}$ Next factor the numerator and denominator. $ r = - \dfrac{1}{n} \cdot \dfrac{(n - 3)(n - 8)}{(n - 3)(n - 3)}$ Assuming $n \neq 3$ , we can cancel the $n - 3$ $ r = - \dfrac{1}{n} \cdot \dfrac{n - 8}{n - 3}$ Therefore: $ r = \dfrac{ -n + 8 }{ n(n - 3)}$, $n \neq 3$